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线性方程组的迭代法求解

发布时间：2024-11-04 10:59:45

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答案：A=[10-1-2;-110-2;-1-15];b=[7.28.34.2]';ep=10^(-5);N=100;x=[];x(:,1)=[000]';x(1,2)=(b(1)-A(1,2)*x(2,1)-A(1,3)*x(3,1))/A(1,1);x(2,2)=(b(2)-A(2,1)*x(1,1)-A(2,3)*x(3,1))/A(2,2);x(3,2)=(b(3)-A(3,1)*x(1,1)-A(3,2)*x(2,1))/A(3,3);k=1;while max(abs(x(:,k+1)-x(:,k)))>ep if k>Nbreak else k=k+1;x(1,k+1)=(b(1)-A(1,2)*x(2,k)-A(1,3)*x(3,k))/A(1,1);x(2,k+1)=(b(2)-A(2,1)*x(1,k)-A(2,3)*x(3,k))/A(2,2);x(3,k+1)=(b(3)-A(3,1)*x(1,k)-A(3,2)*x(2,k))/A(3,3);endendxdisp(A\b)结果:ans =0000.72000.83000.84000.97101.07001.15001.05701.15711.24821.08541.18531.28281.09511.19511.29411.09831.19831.29801.09941.19941.29931.09981.19981.29981.09991.19991.29991.10001.20001.30001.10001.20001.30001.10001.20001.3000方程组的精确解为:1.10001.20001.30002.clearclcA=[10-1-2;-110-2;-1-15];b=[7.28.34.2]';ep=10^(-5);N=100;x=[];x(:,1)=[000]';x(1,2)=(b(1)-A(1,2)*x(2,1)-A(1,3)*x(3,1))/A(1,1);x(2,2)=(b(2)-A(2,1)*x(1,2)-A(2,3)*x(3,1))/A(2,2);x(3,2)=(b(3)-A(3,1)*x(1,2)-A(3,2)*x(2,2))/A(3,3);k=1;while max(abs(x(:,k+1)-x(:,k)))>ep if k>Nbreak else k=k+1;x(1,k+1)=(b(1)-A(1,2)*x(2,k)-A(1,3)*x(3,k))/A(1,1);x(2,k+1)=(b(2)-A(2,1)*x(1,k+1)-A(2,3)*x(3,k))/A(2,2);x(3,k+1)=(b(3)-A(3,1)*x(1,k+1)-A(3,2)*x(2,k+1))/A(3,3);endendx'disp(A\b)结果:ans =0000.72000.90201.16441.04311.16721.28211.09311.19571.29781.09911.19951.29971.09991.19991.30001.10001.20001.30001.10001.20001.30001.10001.20001.3000方程组的精确解为:1.10001.20001.3000

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